Question

(Q2) Calculating the Distribution of Values 10 marks For a set of n real numbers {xi,..., x.J, here are a number of basic statistics we would like to know about the distribution of these numbers: ·The minimum value m = min(x 1, . .., Xn The maximum value Mmax{xi,..., x,J The average value a -^ 1x The standard deviation sx,-a)2. Write a program stats.c that calculates these values for a sequence of values it reads from stdin. The input is provided as a sequence of n + 1 numbers, each on a separate line. The first number is n. The remaining n numbers are xj,..., x,. Your program may assume that n < 1000. For two bonus marks, your code should be able to handle any value of n, even values greater than 1000. For any given input, your program should print the output 2 to stdout, where the are to be replaced with the values of m, M, a, and s. Each output value should be represented with 3 digits after the decimal point. To generate some test data for your program, you can use the genseq program you created earlier. The invocation ./genseq <n> generates a sequence of n random numbers and prints n followed by these n random numbers to stdout. Using ./genseq <n> > seq.txt, you can save this output in a file seq.txt. You can use this as the test input for your stats program by running ./stats < seq.txt Add your file stats.c to SVN and commit it.

Answer 1

C code:

#include <stdio.h>

#include <math.h>

int main()

{

int n,i;

scanf("%d", &n); //read input n

  

float data[n]; //array of size n, to store data

for( i=0; i < n; i++) //to read n number of inputs

{

scanf("%f", &data[i]);

}

  

//minimum value

float minValue = data[0]; //initial assumption

for( i=1; i<n; i++)

{

if (minValue > data[i])

{  

minValue = data[i]; // update with minimum value

}

}

printf("The Minimum value m = %.6f \n", minValue );

  

//maximum value

float maxValue = data[0]; //initial assumption

for( i=1; i<n; i++)

{

if (maxValue < data[i])

{  

maxValue = data[i]; // update with maximum value

}

}   

printf("The Maximum value M = %.6f \n", maxValue );

  

//average

float sum = 0.0, avg;

for(i=0; i<n; i++)

{

sum += data[i]; //sum of all numbers

}

avg = sum/n; //average

printf("The Average value a = %.6f \n", avg );

//Standard Deviation

float sumSquares = 0.0, standardDeviation;

for(i=0; i<n; i++)

{

sumSquares += pow(data[i] - avg, 2); //sum of squares of difference of numbers from average

}

standardDeviation = sqrt(sumSquares/n);

printf("The Standard Deviation s = %.6f \n", standardDeviation);

return 0;

}

Code Screenshot :

CODE OUTPUT | |#1nclude<stdio.h> | #include<math.h> 4 int main int n,i; 7 | scanf("%d". &n); //read input n 9 float data[n)://array of size n, to store data 10 for( i 0; i< ni++) //to read n number of inputs 121 scanf("%f",&data[i]); 14 16 float minValue -data[0;//initial assumption 18 | for( i#1 ; ien i++) //minimum value 20 if (minValue> datalil) 21 minValue datali);//update with minimum value 23 24 25 | printf("The Minimum value m-%.6f\n", minValue ); 26 27 //maximum value 28 float maxValue-data[O; //initial assumption 29 30 for( i-1; i<n; i++) 31 32 if (maxValue < datalil) maxValue-datalil:// update with maximum value 34 35 36 37 | printf("The Maximum value M-%6f \n", maxValue ); 38 39 /average 40 float sum-0.0, avg 41 42 | for(i:0; i«nit+) sum +-data叱//sum of all numbers 47 avg sum/n;//average 48 | printf("The Average value a-%of 49 50 //Standard Deviation 51 float sumSquares 0.0, standardDeviation; 52 for(-0: ini++) .6f \n", avg ); s+pow(datalil - avg, 2); //sum of squares sumSq of difference of numbers from average uare 54 56 standardDeviation sqrt(sumSquares/n 57 | printf("The Standard Deviation s-%.6f\n", standardDeviation); 591 return 60

Output :

CODE OUTPUT The Minimum value m= 1.000000 The Maximum value M = 10.000000 The Average value a = 5.500000 The Standard Deviation s = 2.872281